Question

Exercitiul 1 Sa se determine termenii de rang 10,15,20 ai progresiei aritmetice :a)-4;-1;2;5;...;b)10,6,2,-2,...;c)1:2;1;3:2;2;....

Answer 1

$\displaystyle a).-4,-1,2,5,... \\ a_1=-4,a_2=-1,a_3=2,a_4=5 \\ r=a_2-a_1\Rightarrow r=-1-(-4) \Rightarrow r=-1+4 \Rightarrow r=3 \\ a_n=a+1+(n-1) \cdot 3 \Rightarrow a_{10}=-4+(10-2) \cdot 3 \Rightarrow \\ \Rightarrow a_{10}=-4+9 \cdot 3 \Rightarrow a_{10}=-4+27 \Rightarrow a_{10}=23 \\ a_{15}=-4+(15-1) \cdot 3 \Rightarrow a_{15}=-4+14 \cdot 3 \Rightarrow a_{15}=-4+42 \Rightarrow \\ \Rightarrow a_{15}=38$
$a_{20}=-4+(20-1) \cdot 3 \Rightarrow a_{20}=-4+19 \cdot 3 \Rightarrow a_{20}=-4+57 \Rightarrow \\ \Rightarrow a_{20} =53$
[tex]\displaystyle b).10,6,2,-2,... \\ a_1=10,a_2=6,a_3=2,a_4=-2 \\ r=a_1-a_2 \Rightarrow r=6-10 \Rightarrow r=-4 \\ a_{10}=10+(10-1) \cdot (-4) \Rightarrow a_{10}=10+9 \cdot (-4) \Rightarrow a_{10}=10-36 \Rightarrow \\ \Rightarrow a_{10}=-26 \\ a_{15}=10+(15-1) \cdot (-4) \Rightarrow a_{15}=10+14 \cdot (-4) \Rightarrow a_{15}= 10-56 \Rightarrow \\ \Rightarrow a_{15}=-46 \\ a_{20}=10+(20-1) \cdot (-4) \Rightarrow a_{20}=10+19 \cdot (-4) \Rightarrow a_{20}=10-76 \Rightarrow  \\ \Rightarrow a_{20}=-66 [/tex]
$\displaystyle c).\frac{1}{2} ,1, \frac{3}{2} ,2,... \\ a_1= \frac{1}{2} ,a_2=1,a_3= \frac{3}{2} ,a_4=2 \\ r=a_2-a_1 \Rightarrow r=1- \frac{1}{2} \Rightarrow r= \frac{2-1}{2} \Rightarrow r= \frac{1}{2} \\ a_{10}= \frac{1}{2}+(10-1) \cdot \frac{1}{2} \Rightarrow a_{10}= \frac{1}{2} +9 \cdot \frac{1}{2} \Rightarrow a_{10}= \frac{1}{2}+ \frac{9}{2} \Rightarrow \\ \Rightarrow a_{10}= \frac{10}{2} \Rightarrow a_{10}=5$
$\displaystyle a_{15}= \frac{1}{2} +(15-1) \cdot \frac{1}{2} \Rightarrow a_{15}= \frac{1}{2} +14 \cdot \frac{1}{2} \Rightarrow a_{15}= \frac{1}{2} + \frac{14}{2} \Rightarrow a_{15}= \frac{15}{2} \\ a_{20}= \frac{1}{2} +(20-1) \cdot \frac{1}{2} \Rightarrow a_{20}= \frac{1}{2} +19 \cdot \frac{1}{2} \Rightarrow a_{20}= \frac{1}{2} + \frac{19}{2} \Rightarrow \\ \Rightarrow a_{20}= \frac{20}{2} \Rightarrow a_{20}=10$